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EET 103. Chapter 5 ( Lecture 1). Single Phase Transformer. Introduction to Transformer. A transformer is a device that changes ac electric energy at one voltage level to ac electric energy at another voltage level through the action of a magnetic field. Introduction to Transformer.
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EET 103 Chapter 5 (Lecture 1) Single Phase Transformer
Introduction to Transformer A transformer is a device that changes ac electric energy at one voltage level to ac electric energy at another voltage level through the action of a magnetic field.
Introduction to Transformer • Transformers are constructed of two coils or more placed around the common freeomagnetic core so that the charging flux developed by one will link to the other. • The coil to which the source is applied is called the primary coil. • The coil to which the load is applied is called the secondary coil.
The most important tasks performed by transformers are: • Changing voltage and current levels in electric power systems. • Matching source and load impedances for maximum power transfer in electronic and control circuitry. • Electrical isolation (isolating one circuit from another or isolating DC while maintaining AC continuity between two circuits).
Mutual Inductance • Mutual inductance exits between coils of the same or different dimensions. • Mutual inductance is a phenomenon basic to the operation of the transformer.
Mutual Inductance • A transformer is constructed of 2 coils placed so that the changing flux developed by one will link the other. • The coil to which the source is applied is called primary • The coil which the load is applied is called secondary.
Ideal Transformer An ideal transformer is a lossless device with an input winding and output winding. a = turns ratio of the transformer
Power in ideal transformer Where q is the angle between voltage and current
Impedance transformation through the transformer The impedance of a device – the ratio of the phasor voltage across it in the phasor current flowing through it
Non-ideal or actual transformer Mutual flux
Losses in the transformer • Copper (I2R) losses: Copper losses are the resistive heating in the primary and secondary windings of the transformer. They are proportional to the square of the current in the windings. • Eddy current losses: Eddy current losses are resistive heating losses in the core of the transformer. They are proportional to the square of the voltage applied to the transformer.
Losses in the transformer • Hysteresis losses: Hysteresis losses are associated with the arrangement of the magnetic domain in the core during each half cycle. They are complex, nonlinear function of the voltage applied to the transformer. • Leakage flux: The fluxes ΦLP and ΦLS which escape the core and pass through only one of the transformer windings are leakage fluxes. These escaped fluxes produce a self inductance in the primary and secondary coils, and the effects of this inductance must be accounted for.
Transformer equivalent circuit Ie Ic Im Ep = primary induced voltage Es = secondary induced voltage Vp = primary terminal voltage Vs = secondary terminal voltage Ip = primary current Is = secondary current Ie = excitation current IM = magnetizing current XM = magnetizing reactance IC = core current RC = core resistance Rp = resistance of primary winding Rs = resistance of the secondary winding Xp = primary leakage reactance Xs = secondary leakage reactance
Dot convention • If the primary voltage is positive at the dotted end of the winding with respect to the undotted end, then the secondary voltage will be positive at the dotted end also. Voltage polarities are the same with respect to the dots on each side of the core. • If the primary current of the transformer flows into the dotted end of the primary winding, the secondary current will flow out of the dotted end of the secondary winding.
Exact equivalent circuit the actual transformer The transformer model referred to primary side The transformer model referred to secondary side
Approximate equivalent circuit the actual transformer The transformer model referred to primary side b. The transformer model referred to secondary side
Exact equivalent circuit of a transformer refer to primary side Ep = primary induced voltage Es = secondary induced voltage Vp = primary terminal voltage Vs = secondary terminal voltage Ip = primary current Is = secondary current Ie = excitation current IM = magnetizing current XM = magnetizing reactance IC = core current RC = core resistance Rp = resistance of primary winding Rs = resistance of the secondary winding Xp = primary leakage reactance Xs = secondary leakage reactance
The Equation: Primary side Secondary side
Exact equivalent circuit of a transformer referred to primary side Exact equivalent circuit of a transformer referred to secondary side
aIp Ip Is/a Is Reqs Reqp jXeqp jXeqs + + + + Reqp=Rp+a2Rs Reqs=Rp / a2+Rs aVs Xeqp=Xp+a2Xs jXM Vp Rc Vp/a Vs jXM/a2 Rc/a2 Xeqs=Xp / a2+Xs - - - - Approximate equivalent circuit of a transformer referred to primary side Approximate equivalent circuit of a transformer referred to secondary side
Example 1 • A single phase power system consists of a 480V 60Hz generator supplying a load Zload=4+j3W through a transmission line ZLine=0.18+j0.24W. Answer the following question about the system. • If the power system is exactly as described below (figure 1(a)), what will be the voltage at the load be? What will the transmission line losses be? • Suppose a 1:10 step-up transformer is placed at the generator end of the transmission line and a 10:1 step down transformer is placed at the load end of the line (figure 1(b)). What will the load voltage be now? What will the transmission line losses be now?
ILine ZLoad=0.18+j0.24 ILoad + IG VLoad V=48000V ZLoad=4+j3 - ILoad ILine T1 T2 ZLine=0.18+j0.24 + IG 1:10 10:1 V=48000V VLoad - Figure 1 (a) ZLoad=4+j3 Figure 1 (b)
Solution 5.1 (a) From figure 1 (a) shows the power system without transformers. Hence IG = ILINE = ILoad. The line current in this system is given by
Therefore the load voltage is and the line losses are
(b) From figure 1 (b) shows the power system with the transformers. To analyze the system, it is necessary to convert it to a common voltage level. This is done in two steps i) Eliminate transformer T2 by referring the load over to the transmission’s line voltage level. ii) Eliminate transformer T1 by referring the transmission line’s elements and the equivalent load at the transmission line’s voltage over to the source side. The value of the load’s impedance when reflected to the transmission system’s voltage is
The total impedance at the transmission line level is now The total impedance at the transmission line level (Zline+Z’load) is now reflected across T1 to the source’s voltage level
Notice that Z’’load = 4+j3 and Z’line=0.0018+j0.0024 . The resulting equivalent circuit is shown below. The generator’s current is a) System with the load referred to the transmission system voltage level b) System with the load and transmission line referred to the generator’s voltage level
Knowing the current IG, we can now work back and find Ilineand ILoad. Working back through T1, we get
Working back through T2 gives It is now possible to answer the questions. The load voltage is given by
the line losses are given by Notice that raising the transmission voltage of the power system reduced transmission losses by a factor of nearly 90. Also, the voltage at the load dropped much less in the system with transformers compared to the system without transformers.
Parameter determination of the transformer • Open circuit test • Provides magnetizing reactance and core loss resistance • Obtain components are connected in parallel
Experiment Setup In the open circuit test, transformer rated voltage is applied to the primary voltage side of the transformer with the secondary side left open. Measurements of power, current, and voltage are made on the primary side. Since the secondary side is open, the input current IOC is equal to the excitation current through the shunt excitation branch. Because this current is very small, about 5% of rated value, the voltage drop across the secondary winding and the winding copper losses are neglected.
Admittance Open circuit Power Factor Open circuit Power Factor Angle Angle of current always lags angle of voltage by q
Short circuit test • Provides combined leakage reactance and winding resistance • Obtain components are connected in series
Experiment Setup In the short circuit test, the secondary side is short circuited and the primaryside is connected to a variable, low voltage source. Measurements of power, current, and voltage are made on the primary side. The applied voltage is adjusted until rated short circuit currents flows in the windings. This voltage is generally much smaller than the rated voltage.
Impedances referred to the primary side Power Factor of the current Angle Power Factor Therefore
The equivalent circuit impedances of a 20-kVA, 8000/240-V, 60-Hz transformer are to be determined. The open circuit test and the short circuit test were performed on the primary side of the transformer and the following data were taken: Find the impedances of the approximate equivalent circuit referred to the primary side and sketch that circuit
Per unit System The per unit value of any quantity is defined as Quantity – may be power, voltage, current or impedance
Two major advantages in using a per unit system • It eliminates the need for conversion of the voltages, currents, and impedances across every transformer in the circuit; thus, there is less chance of computational errors. • The need to transform from three phase to single phase equivalents circuits, and vise versa, is avoided with the per unit quantities; hence, there is less confusion in handling and manipulating the various parameters in three phase system.
Voltage Regulation (VR) The voltage regulation of a transformer is defined as the change in the magnitude of the secondary voltage as the current changes from full load to no load with the primary held fixed. At no load,
Phasor Diagram Lagging power factor Unity power factor
Efficiency The efficiency of a transformer is defined as the ratio of the power output (Pout) to the power input (Pin). Pcore = Peddy current + Physteresis And Pcu= Pcopper losses
Pcu= Copper losses are resistive losses in the primary and secondary winding of the transformer core. They are modeled by placing a resistor Rpin the primary circuit of the transformer and resistor Rs in the secondary circuit. PCORE = Core loss is resistive loss in the primary winding of the transformer core. It can be modeled by placing a resistor Rcin the primary circuit of the transformer.
Example A 15 kVA, 2400/240-V transformer is to be tested to determine its excitation branch components, its series impedances and its voltage regulation. The following test data have been taken from the primary side of the transformer
The data have been taken by using the connections of open circuit test and short circuit test Find the equivalent circuit of this transformer referred to the high voltage side. Find the equivalent circuit of this transformer referred to the low voltage side. Calculate the full load voltage regulation at 0.8 lagging power factor and 0.8 leading power factor. What is the efficiency of the transformer at full load with a power factor of 0.8 lagging?